Problem: What is the slope of the line that is tangent to a circle at point (5,5) if the center of the circle is (3,2)? Express your answer as a common fraction.
Answer: If a line can be drawn tangent to a circle a the point $(5,5)$, then it must be possible to draw a radius from the center of the circle to the point $(5,5)$.  This radius will have a slope of: $$\frac{5-2}{5-3}=\frac{3}{2}$$ A key fact to remember is that tangents to a circle at a certain point are perpendicular to radii drawn from the center of the circle to that point.  This diagram summarizes that fact: [asy]
draw(Circle((0,0),sqrt(13)),linewidth(.8));
draw((-1,5)--(5,1),linewidth(.8));
draw((0,0)--(2,3),linewidth(.8));
draw((2-0.3,3+0.2)--(2-0.5,3-0.1)--(2-0.2,3-0.3));
[/asy] Therefore, the slope of the tangent will be the negative inverse of the slope of the radius, which is equal to $\boxed{-\frac{2}{3}}$.